Using proof by contradiction for this Pigeonhole Principle problem
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Using proof by contradiction for this Pigeonhole Principle problem

[From: ] [author: ] [Date: 11-12-06] [Hit: ]
I just want to know if I recalled the proof correctly.Thank you.-On the right lines. Heres how I would put it,Assume each child received at most 4 pieces of candy. Then the number of pieces of candy received must be at most 15*4 = 60.......
Suppose that at a competition 15 children received 63 pieces of candy. Show that at least one child received at least 5 pieces of candy.

My prof told me that I could use proof by contradiction, but I forgot what she did. She said something like:

Assume, to the contrary, that each child receives at most 4 pieces of candy. Then ceiling(63/15) = 5, a contradiction.

Is this correct? I just want to know if I recalled the proof correctly.

Thank you.

-
On the right lines. Here's how I would put it, not even using the ceiling function:

Assume each child received at most 4 pieces of candy. Then the number of pieces of candy received must be at most 15*4 = 60. But there were 63 pieces of candy received - a contradiction. Hence at least one child received at least 5 pieces of candy.

Your proof is correct, but it seems to be that there is a little bit missing between the assumption and your use of the ceiling that would make it clearer.
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